Question

solve for x. figures are not necessarily drawn to scale.

Explanation:

Step1: Identify Similar Triangles

Since $\angle DFE = \angle C = 63^\circ$ and $\angle D$ is common, $\triangle DFE \sim \triangle DCA$ by AA similarity. So, the ratios of corresponding sides are equal: $\frac{DE}{DA}=\frac{DF}{DC}$.
$DA = DE + EA = 24 + 44 = 68$? Wait, no, wait: Wait, $DE$ is 24? Wait, no, the segment from D to E is 24, and E to B is 44? Wait, no, looking at the diagram: $DB$ is composed of $DE = 24$ and $EB = 44$? Wait, no, the left side: from D to E is 24, E to B is 44? Wait, no, the label "24" is on DE, "44" on EB? Wait, no, the curved bracket on the left: D to E is 24, E to B is 44? Wait, no, the total from D to B would be 24 + 44 = 68? Wait, no, maybe the sides: $DE = 24$, $EB = 44$, so $DB = 24 + 44 = 68$? Wait, no, the right side: $DC$ is composed of $DF = x$ and $FC = 49.5$, so $DC = x + 49.5$. And since $\triangle DFE \sim \triangle DCA$, the ratio of $DE$ to $DB$? Wait, no, wait: Wait, $E$ is on $DB$ and $F$ is on $DC$, and $EF \parallel BC$ (because $\angle DFE = \angle C = 63^\circ$, so corresponding angles, so $EF \parallel BC$). Therefore, by the Basic Proportionality Theorem (Thales' theorem), $\frac{DE}{DB}=\frac{DF}{DC}$. Wait, $DE = 24$, $DB = DE + EB = 24 + 44 = 68$? Wait, no, maybe $DE = 24$, $EB = 44$, so $DB = 24 + 44 = 68$? Wait, no, the problem: Let's re-express. Let $DE = 24$, $EB = 44$, so $DB = 24 + 44 = 68$. $DF = x$, $FC = 49.5$, so $DC = x + 49.5$. Since $EF \parallel BC$ (because $\angle DFE = \angle C = 63^\circ$, so alternate interior angles? Wait, $\angle DFE$ and $\angle C$ are both $63^\circ$, so $EF \parallel BC$, so by the Basic Proportionality Theorem (Thales' theorem), $\frac{DE}{DB}=\frac{DF}{DC}$. Wait, no, Thales' theorem: if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. So in $\triangle DBC$, $EF \parallel BC$, so $\frac{DE}{DB}=\frac{DF}{DC}$. Wait, $DE = 24$, $DB = DE + EB = 24 + 44 = 68$? Wait, no, maybe $DE = 24$, $EB = 44$, so $DB = 24 + 44 = 68$? Wait, no, maybe the labels are different. Wait, maybe $DE = 24$, $DB = 44$? No, the curved bracket on the left: D to E is 24, E to B is 44? So $DB = 24 + 44 = 68$. And $DC = DF + FC = x + 49.5$. Then by Thales' theorem: $\frac{DE}{DB}=\frac{DF}{DC}$. So $\frac{24}{24 + 44}=\frac{x}{x + 49.5}$. Wait, $24 + 44 = 68$, so $\frac{24}{68}=\frac{x}{x + 49.5}$. Cross-multiplying: $24(x + 49.5) = 68x$. $24x + 24 \times 49.5 = 68x$. $24 \times 49.5 = 68x - 24x = 44x$. $24 \times 49.5 = 1188$. Then $44x = 1188$. $x = \frac{1188}{44} = 27$. Wait, that makes sense. Let's check: $24/68 = 6/17$, and $27/(27 + 49.5) = 27/76.5 = 270/765 = 6/17$. Yes, that works. So the steps:

Step1: Identify Similar Triangles

Since $\angle DFE = \angle C = 63^\circ$ and $\angle D$ is common, $\triangle DFE \sim \triangle DCA$ (AA similarity). Alternatively, since $EF \parallel BC$ (corresponding angles equal), by Thales' theorem, $\frac{DE}{DB}=\frac{DF}{DC}$.

Step2: Define Segments

Let $DE = 24$, $EB = 44$, so $DB = DE + EB = 24 + 44 = 68$. Let $DF = x$, $FC = 49.5$, so $DC = DF + FC = x + 49.5$.

Step3: Set Up Proportion

By Thales' theorem: $\frac{DE}{DB}=\frac{DF}{DC}$
Substitute values: $\frac{24}{68}=\frac{x}{x + 49.5}$

Step4: Solve for $x$

Cross - multiply: $24(x + 49.5)=68x$
Expand: $24x + 24\times49.5 = 68x$
Calculate $24\times49.5 = 1188$: $24x + 1188 = 68x$
Subtract $24x$: $1188 = 44x$
Divide: $x=\frac{1188}{44}=27$

Answer:

$27$